点击消除 题解
点击消除
https://ac.nowcoder.com/acm/contest/5600/E
这里我是用了一个num数组来记录对应位置的字符是否保留。
然后往后找找到两个相同的字符，然后用双指针往两边展开，跳过已经被删除的字符。
最后只需要把还保留的字符输出即可
import java.util.*;
import java.math.*;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.io.OutputStreamWriter;
import java.io.BufferedReader;
import java.io.PrintWriter;
public class Main {
public static HashMap<Integer,Long>map = new HashMap<>();
public static void main(String args[])throws IOException {
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
in.nextToken();
String s = in.sval;
int num[] = new int[s.length()];
Arrays.fill(num,1);
int l=0,r=0;
String s1="";
for(int i=1;i<s.length();i++)
{
if(num[i]!=0&&num[i-1]!=0&&s.charAt(i)==s.charAt(i-1))
{
r = i;
l = i-1;
while(l>=0&&r<s.length()&&s.charAt(l)==s.charAt(r))
{
num[l] = 0;
num[r] = 0;
l--;r++;
while(l>=0&&num[l]==0)
l--;
while(r<s.length()&&num[r]==0)
r++;
}
}
}
for(int i=0;i<s.length();i++)
{
if(num[i]!=0)
s1+=s.charAt(i);
}
if(s1.length()==0)
out.print(0);
else{
out.print(s1);
}
out.flush();
}
} 
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